We know $f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)(x-r_6)$, where $r_k$ is the $k$th root. This means the term with no $x$’s can only be obtained by multiplying (the negative of) all the roots together (this is a well known result, as per Vieta’s Formulas). Then $64 = r_1 r_2 r_3 r_4 r_5 r_6$. By a similar argument, $e = r_1 r_2 r_3 r_4 r_5 + \cdots + r_2 r_3 r_4 r_5 r_6$.